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#11
11-03-2014, 11:18 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477
Re: Discussion of the VC proof

Quote:
 Originally Posted by jokovc Hi Prof. Yaser. I have a problem with the proof of Lemma A.2., page 190. I don't understand what this part means
Here is what it means. Let's say that you have for every . Then, regardless of what the probability distribution of , it will be true that since we can multiply both sides of by and integrate out.
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#12
11-04-2014, 08:24 PM
 jokovc Junior Member Join Date: Nov 2014 Posts: 2
Re: Discussion of the VC proof

Quote:
 Originally Posted by yaser Here is what it means. Let's say that you have for every . Then, regardless of what the probability distribution of , it will be true that since we can multiply both sides of by and integrate out.
I think this is so much clearer than when it is put in sentences. Thank you.
#13
10-01-2015, 07:32 PM
 ilson Member Join Date: Sep 2015 Posts: 10
Re: Discussion of the VC proof

I was going through the proof in Appendix A and I just want to make sure that something written towards the bottom of pg. 189 is a typo.

Namely,

Quote:
 Inequality (A.3) folows because the events "" and "" (which is given) imply "".

Quote:
 Inequality (A.3) folows because the events "" and "" (which is given) imply "".
I'm 99.999999999999% sure this is indeed a typo since the latter case easily follows from reverse triangle inequality and it suffices to show the inequality in (A.3) and I cannot see how one can arrive at the implication in the former case nor how the former case implies the inequality (A.3), but it would ease my mind if I can get a verification that it is a typo. Thank you in advance!
#14
10-13-2015, 12:04 PM
 magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 595
Re: Discussion of the VC proof

Yes, this is a typo. Thank you for pointing it out. You have it correct.

If A and C imply B, then

Quote:
 Originally Posted by ilson I was going through the proof in Appendix A and I just want to make sure that something written towards the bottom of pg. 189 is a typo. Namely, should read I'm 99.999999999999% sure this is indeed a typo since the latter case easily follows from reverse triangle inequality and it suffices to show the inequality in (A.3) and I cannot see how one can arrive at the implication in the former case nor how the former case implies the inequality (A.3), but it would ease my mind if I can get a verification that it is a typo. Thank you in advance!
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#15
09-26-2016, 09:56 PM
 CharlesL Junior Member Join Date: Sep 2016 Location: Vancouver Posts: 1
Re: Discussion of the VC proof

Recently I have started reading the proof of the VC inequality in the appendix. On the bottom of page 190(Lemma A.3) , why does

sigma_S　P[S] x P[sup_h E_in - E_in' > ... |S ] <= sup_S P[sup_h E_in - E_in' >．．．|S ]?
(Sorry for the terrible notations, I don't know how I can input math symbols)

what does it mean by taking the supremum on S?
#16
10-05-2016, 02:23 PM
 htlin NTU Join Date: Aug 2009 Location: Taipei, Taiwan Posts: 601
Re: Discussion of the VC proof

Quote:
 Originally Posted by CharlesL Recently I have started reading the proof of the VC inequality in the appendix. On the bottom of page 190(Lemma A.3) , why does sigma_S　P[S] x P[sup_h E_in - E_in' > ... |S ] <= sup_S P[sup_h E_in - E_in' >．．．|S ]? (Sorry for the terrible notations, I don't know how I can input math symbols) what does it mean by taking the supremum on S?
All complicated math aside, supremum on S carries the physical meaning of taking the "maximum" value over all possible S.

So this inequality simply says an expected value (of P[sup_...]) is less than or equal to the maximum value.

Hope this helps.
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#17
10-08-2016, 03:44 PM
 tayfun29 Junior Member Join Date: Oct 2016 Posts: 1
Re: Discussion of the VC proof

Quote:
 Originally Posted by htlin All complicated math aside, supremum on S carries the physical meaning of taking the "maximum" value over all possible S. So this inequality simply says an expected value (of P[sup_...]) is less than or equal to the maximum value. Hope this helps.
Thank...
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#18
10-25-2016, 07:39 AM
 gamelover623 Junior Member Join Date: Sep 2016 Posts: 1
Re: Discussion of the VC proof

I am a machine learning practitioner currently applying machine learning to algorithmic trading, yet highly interested in the theoretical grounds of the field.

I have read your book "Learning from data" from cover to cover. I haven't solved the problems though. I however did go through the proof of the VC bound in the appendix. I succeeded to understand most of it except (A.4) in the bottom of page 189. I can understand that you have applied Hoeffding Inequality to h*, but your explanation on how this applies to h* conditioned to the sup_H event, is hard to grasp for me.

Can you please give more explanation on how using Hoeffding (A.4) holds ? Or give a reference that helps clarifying this result?

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#19
10-26-2016, 03:23 AM
 CountVonCount Member Join Date: Oct 2016 Posts: 17
Re: Discussion of the VC proof

Hi,

I have a question about the sentence on page 190:
Quote:
 Note that we can assume e^(-0.5*N*eps^2) < 1/4, because otherwise the bound in Theorem A.1 is trivially true.
While I understand the argument here, I don't understand, why it is especially the value 1/4?
When set the above term to 1/4 I will receive -2*ln(1/4) as value for N*eps^2.
Now I can set N*eps^2 to that value in Theorem A.1 and I will get on the RHS (assuming the growth function is just 1) 4*0,707... so it is much more than 1.

A value of 1 in the RHS would be sufficient to say the bound in Theorem A.1 is trivially true. And this would assume, that the above term is less than 1/256.
With this in mind 1 - 2*e^(-0.5*N*eps^2) is greater than 0,99... and thus instead of a 2 in the lemmas outcome, I would receive a value around 1, which is a much better outcome.

So why is the value 1/4 chosen for the assumption?

Best regards,
André
#20
10-26-2016, 01:38 PM
 magdon RPI Join Date: Aug 2009 Location: Troy, NY, USA. Posts: 595
Re: Discussion of the VC proof

Suppose

Then, .

In which case and the bound in Theorem A.1 is trivial.

Quote:
 Originally Posted by CountVonCount Hi, I have a question about the sentence on page 190: While I understand the argument here, I don't understand, why it is especially the value 1/4? When set the above term to 1/4 I will receive -2*ln(1/4) as value for N*eps^2. Now I can set N*eps^2 to that value in Theorem A.1 and I will get on the RHS (assuming the growth function is just 1) 4*0,707... so it is much more than 1. A value of 1 in the RHS would be sufficient to say the bound in Theorem A.1 is trivially true. And this would assume, that the above term is less than 1/256. With this in mind 1 - 2*e^(-0.5*N*eps^2) is greater than 0,99... and thus instead of a 2 in the lemmas outcome, I would receive a value around 1, which is a much better outcome. So why is the value 1/4 chosen for the assumption? Best regards, André
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